Author Topic: VCE Methods Question Thread! (Read 5783163 times) Tweet Share

brightsky · « **Reply #3810 on:** January 21, 2014, 04:35:54 pm »

No, you are absolutely correct in thinking that the double angle formula ought to be used.

soNasty · « **Reply #3811 on:** January 21, 2014, 04:38:08 pm »

great, thank you!

alchemy · « **Reply #3812 on:** January 21, 2014, 04:42:05 pm »

Quote

Could the answer have also been $a = \frac{100}{10^{b}}$ ??
Does it equal the same thing or am I just doing illegal math

Yes, that's correct

b^3 · « **Reply #3813 on:** January 21, 2014, 05:04:28 pm »

Well looking back at it I screwed up that last line and screwed up a negative earlier as well, oddly enough it doesn't actually affect the result BUT there is a third thing I forgot to mention. When we square both sides, we sometimes are introducing new solutions that aren't a solution to the original equation. So take for example if we have $\sqrt{x}=1-x$ , when we square both sides we are assuming that $1-x$ is positive, because we know that our LHS has to be positive because of the square root, and so $1-x\geq0$ . If we don't account for this restriction then we may get another solution that is the solution to $x=(1-x)^{2}$ but not the solution to $\sqrt{x}=1-x$ .

I'll go and fix it on the page before, but this should be the new working (plus the old restrictions as well)
$\begin{alignedat}{1}\sqrt{5-x}-\sqrt{x} & =\sqrt{2x-1} \\ \left(\sqrt{5-x}-\sqrt{x}\right)^{2} & =2x-1 \\ 5-x-2\sqrt{5-x}\sqrt{x}+x & =2x-1 \\ -2\sqrt{5-x}\sqrt{x} & =2x-6 \\ \sqrt{5x-x^{2}} & =3-x \\ 3-x\geq0 \text{ as }\sqrt{5x-x^{2}}\geq0 \\ x\leq3 \\ 5x-x^{2} & =\left(3-x\right)^{2} \\ 5x-x^{2} & =9-6x+x^{2} \\ 2x^{2}-11x+9 & =0 \\ \left(2x-9\right)\left(x-1\right) & =0 \\ x=\frac{9}{2},\: x=1 \\ \text{But }x\leq3 \\ \therefore x=1 \end{alignedat}$

soNasty · « **Reply #3814 on:** January 21, 2014, 05:06:18 pm »

sorry can someone help me simplify

$\sqrt{3}+1/1-\sqrt{3}$

not sure what the conjugate denominator would be

b^3 · « **Reply #3815 on:** January 21, 2014, 05:14:18 pm »

Quote from: andrew2910 on January 21, 2014, 05:06:18 pm

sorry can someone help me simplify

$\sqrt{3}+1/1-\sqrt{3}$

not sure what the conjugate denominator would be

Denominator is $1-\sqrt{3}$ so multiply by $\frac{1+\sqrt{3}}{1+\sqrt{3}}$ .

Phy124 · « **Reply #3816 on:** January 21, 2014, 05:15:13 pm »

chucking this in a spoiler 'cause b^3 already told you how to go about it

$\frac{\sqrt{3}+1}{1-\sqrt{3}} = \frac{\sqrt{3}+1}{1-\sqrt{3}}\times\frac{1+\sqrt{3}}{1+\sqrt{3}} = \frac{\sqrt{3}+3+1+\sqrt{3}}{1-3}=-\frac{4+\sqrt{3}}{2}=-\sqrt{3}-2$

alchemy · « **Reply #3817 on:** January 21, 2014, 06:23:03 pm »

Determine the exact value for $x$ :
$\log_2 4x+\log_4 8x+\log_8 2x = 4$

Suggested solution:
$\log_2 4x+\log_4 8x+\log_8 2x = 4 \log_{64}4^6x^6+\log_{64}8^3x^3+\log_{64}2^2x^2 = 4 \log_{64}2^{12+9+2}x^{6+3+2}=4 2^{23}x^{11}=2^{24} x^{11}=2 x=\sqrt[11]{2}$

I don't understand the suggested solution as I don't do MM 3/4 yet. Can someone please explain the above OR an alternative easier method?

IndefatigableLover · « **Reply #3818 on:** January 22, 2014, 12:44:03 am »

Alright so for both questions you don't need a calculator for (I did this without a calculator)

a). So from $sin( \frac{2 \pi x}{3})=- \frac{\sqrt3}{2}}$ we find the basic angle for this.

Spoiler

$sin\theta=-\frac{\sqrt3}{2}}$

$\Theta = \frac{\pi}{3}$

Because $-\frac{\sqrt3}{2}}$ is negative, we know that for sine, it is negative in the 3rd and 4th quadrant.

From here we go around the unit circle to the 3rd quadrant to $\pi$ and add $\frac{\pi}{3}$ which gives us $\frac{4\pi}{3}$
Likewise, we can subtract $2\pi$ from $\frac{\pi}{3}$ which will give us $\frac{5\pi}{3}$

Now we know that $\theta = \frac{2 \pi x}{3}$ so:

$\frac{2 \pi x}{3} = \frac{5\pi}{3}$
$\frac{2 \pi x}{3} = \frac{4\pi}{3}$

${2 \pi x} = {5\pi}$
${2 \pi x} ={4\pi}$

$x = \frac{5}{2}$
$x = 2$

And both answers are in your restriction of $x \in [0,3]$ so your answers are $x = \frac{5}{2} and x = 2$

Will work on the next one when I wake up (such a late night LOL [hope this is right :/

brightsky · « **Reply #3819 on:** January 22, 2014, 12:55:44 am »

Let $f:R \rightarrow R, f(x) = sin( \frac{2 \pi x}{3})$ .

a. Solve the equation $sin( \frac{2 \pi x}{3}) = - \frac{\sqrt3}{2}} for x \in [0,3]$

since x E [0,3], therefore 2pi/3 x E [0, 2pi]

sin(2pi/3 x) = -sqrt(3)/2
2pi/3 x = pi + pi/3, 2pi - pi/3 (only choose solutions in the domain specified above)
2pi/3 x = 4pi/3, 5pi/3
x = 2, 5/2

different teachers will teach different methods to solve trig equations. choose a method that you're comfortable with and stick to that method.

b. Let $g:R \rightarrow R, g(x) = 3f(x-1)+2$ . Find the smallest positive value of x for which g(x) is a maximum.

g(x) = 3 sin(2pi/3*(x-1)) + 2

now since sin(blah) is between -1 and 1 (inclusive), we know that the range of g(x) is [-1,5]. so the maximum value of g(x) is 5. now when is this true? when sin(2pi/3*(x-1)) = 1. so we solve:

sin(2pi/3*(x-1)) = 1
2pi/3*(x-1) = ..., -7pi/2, -3pi/2, pi/2, 5pi/2, 9pi/2, ...
2/3*(x-1) = ..., -7/2, -3/2, 1/2, 5/2, 9/2, ...
x-1 = ..., -21/4, -9/4, 3/4, 15/4, 27/4, ...
x = ..., -17/4, -5/4, 7/4, 19/4, 31/4, ...

we require the smallest positive solution for x. so x = 7/4.

hopefully no errors.

scribble · « **Reply #3820 on:** January 22, 2014, 10:28:50 am »

Quote from: alchemy on January 21, 2014, 06:23:03 pm

Determine the exact value for $x$ :
$\log_2 4x+\log_4 8x+\log_8 2x = 4$

Suggested solution:
$\log_2 4x+\log_4 8x+\log_8 2x = 4 \log_{64}4^6x^6+\log_{64}8^3x^3+\log_{64}2^2x^2 = 4 \log_{64}2^{12+9+2}x^{6+3+2}=4 2^{23}x^{11}=2^{24} x^{11}=2 x=\sqrt[11]{2}$

I don't understand the suggested solution as I don't do MM 3/4 yet. Can someone please explain the above OR an alternative easier method?

the reason why this is hard to solve is because we have three separate logs. if we can combine them into just one log, it would make life much easier. Now you'll learn a couple of log laws that allow you to combine logs, but these can only be applied when the logs all have the same base. so for this question, we need to convert the three logs so that they all share the same base system. so how can we do that? first you need to think about what a log is, and how it works.
if i have
$\ 2^4=y$ i can write it out as $\ log_2(y)=4$ right?
now if i take the first equation, and square both sides, i get
$\ (2^4)^2=y^2$ which is the same as $\ (2^2)^4=y^2$
i can write this as a log again:
$\ log_{2^2}(y^2)=4$ or $\ log_4(y^2)=4$
so you can see that if I square the base of the log, i also need to square whats inside the log.

now in the question, we have 2, 4 and 8 as bases.
if we take the first, second, third,... power of 2, we get
2,4,8,16,32,64 etc
likewise for 4, we have
4,16,64, 256
and for 8, we have
8,64,512
you'll see that 64 is common for these three bases, so we want to convert our three logs, so that they all have the base 64.
$2^6 = 64$ so $\log_2 4x$ becomes $log_{2^6}((4x)^6)$ which is equal to $log_{64}(4^6x^6)$
likewise, $log_4 8x = log_{4^3}((8x)^3) = log_{64}(8^3x^3)$
and $log_8(2x) = log_{64}(2^2x^2)$

so now the equation is
$log_{64}(4^6x^6) + log_{64}(8^3x^3) + log_{64}(2^2x^2) = 4$
and from log laws, we can combine the three logs into one by multiplying the arguments, so
$log_{64}(4^6x^6) + log_{64}(8^3x^3) + log_{64}(2^2x^2) = log_{64}(4^6x^6 * 8^3x^3 * 2^2x^2)$
and using exponential rules, we can simplify this further
$log_{64}(4^6x^6 * 8^3x^3 * 2^2x^2) = log_{64}(2^{12}x^6*2^9x^3*2^2x^2 = log_{64}(2^{23}x^{11})$
so now we just have
$log_{64}(2^{23}x^{11}) = 4$ as our equation, and we can easily solve this;
rearranging to get rid of the log, we have
$64^4 = 2^{23}x^{11}$
$2^{6*4} = 2^{23}x^{11}$
$2^{24}=2^{23}x^{11}$
$2=x^{11}$
$x=2^{1/11}$
which is your final answer

liamh96 · « **Reply #3821 on:** January 22, 2014, 03:06:23 pm »

Hi guys,

I've been asked to find f(h(x)) and h(f(x))
Where f(x)= { x^2 -4 , x belongs to (-infinity, 2)
{ x , x belongs to [2, infinity)
and h(x)= 2x
Does anyone know how to solve this question?

Jason12 · « **Reply #3822 on:** January 22, 2014, 03:38:20 pm »

Find the equation of the straight line which passes through the point (1,6) and is parallel to the line with the equation 4x+2y=10 and perpendicular to the line with equation 4x+2y=10

Conic · « **Reply #3823 on:** January 22, 2014, 04:09:25 pm »

Quote from: Jason12 on January 22, 2014, 03:38:20 pm

Find the equation of the straight line which passes through the point (1,6) and is parallel to the line with the equation 4x+2y=10 and perpendicular to the line with equation 4x+2y=10

Since the line is parallel to 4x+2y=10, it will have the same gradient.

$4x+2y=10 \Rightarrow 2x+y=5$ (divide by 2)

$2x+y=5 \Rightarrow y=5-2x$ (take 2x from both sides)

The gradient of the line we are trying to find is -2, so the equation of the line is -2x+c. Now we use the point (1,6) to find c.

$6=-2 \times 1+c \Rightarrow c=8$ (rearrange for c)

So the line is y=-2x+8, or 2x+y=8.

Stick · « **Reply #3824 on:** January 22, 2014, 04:21:06 pm »

Quote from: liamh96 on January 22, 2014, 03:06:23 pm

Hi guys,

I've been asked to find f(h(x)) and h(f(x))
Where f(x)= { x^2 -4 , x belongs to (-infinity, 2)
{ x , x belongs to [2, infinity)
and h(x)= 2x
Does anyone know how to solve this question?

Alright, so this question is going to get a little bit involved. It's going to be fairly straightforward to find the equations, but the difficulty associated with this particular problem lies in the domains and ranges. I'll do the first bit for you (we'll find the rule $f \circ h(x)$ for the left branch) and hopefully you can use that to try and work out and practice the other parts.

Let's just deal with the left branch of the hybrid function $f(x)$ (we'll call this $f_1(x)$ for the sake of simplicity) and the function $h(x)$ .

So $f_1(x)=x^2-4, x \in (- \infty , 2)$
and $h(x)=2x, x \in R$

Now, the rule attached with composite functions is that the range of the function in the "inside" must be a subset of the domain of the function on the "outside." This may also be more condensely expressed as "the range of the inside must be a subset of the domain of the outside" and you may find the term "R.I.D.O." helpful to remember this (range inside - subset of - domain outside). I happened to remember this because it was the nickname of my old deputy principal.

Anyway, let's take a closer look at the domains and ranges:

$f_1(x)$ : Domain $(-\infty,2)$ Range $[-4,\infty)$
$h(x)$ : Domain $R$ Range $R$

For our composite function to exist, we need the range of $h(x)$ to be a subset of (or "fit inside") the domain of $f_1(x)$ . At the moment, this isn't the case. Therefore, we need to restrict the range of $h(x)$ to $(-\infty,2)$ . To do that, the domain of $h(x)$ must be changed to $(-\infty,1)$ . Now that we've made that adjustment, everything works nicely.

It's time for us to obtain our final result by simply inserting the inside function into the outside function. The domain of our new composite function is the domain of the inside function. This must be stated every time we create a composite function.

Hence, $f_1 \circ h(x)=(2x)^2-4$
$f_1 \circ h(x)=4x^2-4, x \in (-\infty, 1)$

Tip: You'll find that you should get hybrid functions once you work out the two composite functions required completely.

I hope this helps.

Search the forums now!

We have moved!

Author Topic: VCE Methods Question Thread! (Read 5783163 times) Tweet Share

brightsky

Re: VCE Methods Question Thread!

soNasty

Re: VCE Methods Question Thread!

alchemy

Re: VCE Methods Question Thread!

b^3

Re: VCE Methods Question Thread!

soNasty

Re: VCE Methods Question Thread!

b^3

Re: VCE Methods Question Thread!

Phy124

Re: VCE Methods Question Thread!

alchemy

Re: VCE Methods Question Thread!

IndefatigableLover

Re: VCE Methods Question Thread!

brightsky

Re: VCE Methods Question Thread!

scribble

Re: VCE Methods Question Thread!

liamh96

Re: VCE Methods Question Thread!

Jason12

Re: VCE Methods Question Thread!

Conic

Re: VCE Methods Question Thread!

Stick

Re: VCE Methods Question Thread!

Recent Posts

User:
Password: